Joint Probability: Explanation, formula and solved examples

Ans. Number of total possible outcomes when a die is rolled = 6

i.e. {1, 2, 3, 4, 5, 6}

Let A be the event of occurring 3 on die 1 and B be the event of occurring 3 on the die 2.

Both the dice have six possible outcomes in all. Therefore, the probability of a three occurring on each die comes out to be 1/6.

P(A) =1/6

P (B) =1/6

P (A ∩ B) = 1/6 x 1/6 = 1/36

Ans. Event A = The probability of rolling a dice and getting 5 in the first roll is 1/6 = 0.1666.

Event B = The probability of rolling a dice and getting 5 in the second roll is 1/6 = 0.1666.

 Therefore, the joint probability of event A and B occurring is P(1/6) x P(1/6) = 0.02777 = 2.8%.

Ans. Event A = The probability of getting a head in the first toss comes out to be 1/2 = 0.5.

Event B = The probability of getting a tail in the second toss comes out to be 1/2 = 0.5.

Therefore, the joint probability of event A and B occurring simultaneously = P (1/2) x P (1/2) = 0.25 = 25%.

Ans. Event A = The probability of drawing a card with number 10 = 4/52 = 0.0769

Event B = The probability of drawing a card which is black = 26/52 = 0.50

Therefore, the joint probability of event A and B occurring together is P (4/52) x P (26/52) = 0.0385 = 3.9%.

Answer: No, both the events do not happen at the synchronously while calculating the conditional probability. One event has already taken place in the past and we are using this inference to find the occurrence of another event that will happen in the near future.

Answer: Yes, there are practical applications of conditional probability in daily life. For instance, consider that weather forecast predicted the probability of rainfall to be 50%. This depends on the information available to them on various factors like wind intensity, temperature, and quantitative analysis of humidity. Therefore, conditional probability can be used in real life.

Answer: Two properties of conditional probability are:

1. Imagine that A and B are the events of sample space which is called S, then the formula would become

P(A|B) = P(B|B) = 1.

2. Another property when represented by using the formula becomes:

P(E′|F) = 1 − P(E|F), where E and F are the two events

Ans. Consider H1 as the event items produced by A, H2 as the event item produced by B and, H3 as the event items produced by C

P(H₁) = \(\frac{50}{100}\), P(H₂) = \(\frac{30}{100}\) and P(H₃) = \(\frac{20}{100}\)

Let E be the event items found to be defective.

P\((\frac{E}{H_1}) = \frac{1}{100}\), P\((\frac{E}{H_2}) = \frac{5}{100}\) and P\((\frac{E}{H_3}) = \frac{7}{100}\)

Using Bayes’ theorem,

P\((\frac{H_1}{E})\) = \(\frac{P(H_1)P(E/H_1)}{P(H_1)P(E/H_1) + P(H_2)P(E/H_2) + P(H_3)P(E/H_3)}\)

= \(\frac{\frac{50}{100} \times \frac{1}{100}}{\frac{50}{100} \times \frac{1}{100} + \frac{30}{100} \times \frac{5}{100} \times \frac{20}{100} \frac{7}{100}}\)

= \(\frac{50}{50+150+140} = \frac{50}{340} = \frac{5}{34}\)

Ans. Given, P(not A) = 0.7

P(B) = 0.7

and,      P(B/A) = 0.5

We know that

P(B/A) = \(\frac{P(A\bigcap B)}{P(A)}\)

0.5 = \(\frac{P(A\bigcap B)}{0.3}\)

[\(\because\) P(not A) = 0.7 then P(A) = 1 – 0.7 = 0.3]

⇒ P(A∩B) = 0.15

Also we know, 

P(A/B) = \(\frac{P(A\bigcap B)}{P(B)}\)

= \(\frac{0.15}{0.7} = \frac{15}{7}\)

Ans.We are given with three coins. Let us consider E1 as the coins with two heads, E2 as the biased coin and E3 as the unbiased coin

Here, P(E₁) = P(E₂) = P(E₃) = \(\frac{1}{3}\)

Then, P\((\frac{A}{E_1}) = 1\)

\((\frac{A}{E_2}) = \frac{75}{100} = \frac{3}{4}\) given

P \((\frac{A}{E_3}) = \frac{1}{2}\)

Now, Probability of two headed coin P(E₁/A)

= \(\frac{P(E_1)P(A/E_1)}{P{E_1) \times P(A/E_1} + P(E_2) \times P(A/E_2) + P(E_3) \times P(A/E_3)} \)

= \(\frac{\frac{1}{3} \times 1}{\frac{1}{3} \times 1 + \frac{1}{3} \times \frac{3}{4} + \frac{1}{3} \times \frac{1}{2}}\) = \(\frac{1}{\frac{4+3+2}{4}} = \frac{4}{9}\)

Ans. The sample space shows 36 outcomes in all. Let us consider A to be the event that gives a sum total of the 8 observations.

Thus, A = {(2,6) (3,5) (5,3) (4,4) (6,2)}

⇒ n(A) = 5

⇒ P(A) = \(\frac{5}{36}\)

Let B be event that observation on red die is less than 4.

∴ B = {1,1), (2,1), (3,1), (4,1), (5,1), (6,1), (1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (1,3), (2,3), (3,3), (4,3), (5,3), (6,3)}

∴ n(B) = 18

⇒ P(A) = \(\frac{18}{36}\) = \(\frac{1}{2}\)

Clearly A∩B = {(5,3), (6,2)}

∴ n(A∩B) = 2

∴ P(A∩B) = \(\frac{2}{36} = \frac{1}{18}\)

P\((\frac{A}{B}) \) = \(\frac{P(A\bigcap B)}{P(B)} = \frac{\frac{1}{18}}{\frac{1}{2}}\)

= \(\frac{2}{18} = \frac{1}{9}\)

Ans. Here, consider E1 as the event in which the girl gets 1 or 2 on the dice roll and E2 as the event that the girl obtains 3, 4, 5, or 6 on the dice role:

Therefore P (E1) = 26= 13

P(E2) = 46= 23

Now let us consider that A is the event in which she obtains exactly one tail. If she tosses the coin thrice and obtains exactly one tail then all the possible outcomes will be HTH, HHT, THH

∴ P(A/E₁) = \(\frac{3}{8}\)

If she tossed coin only once and exactly one tail shows 

Then, P(A/E₂) = \(\frac{1}{2}\)

∴ P(E₂/A) = \(\frac{P(E_2).P(A/E_2)}{P(E_1).P(A/E_1) + P(E_2).P(A/E_2)}\)

= \(\frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}}\) = \(\frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{11}{24}}\)

= \(\frac{8}{11}\)

Ans. P (one red and one black) = P (first red and second black) + P (first black and second red)

= 26/ 52 x 26/ 51 + 26/ 52 x 26/ 51 [ without replacement]

= 13/ 51 + 13/ 51

= 26/ 51

Ans. The repeated tosses of a die are known as Bernoulli trials. If the number of successes of getting odd numbers in an experiment of 6 trials is considered to be X, then

The probability of getting an odd number in a single throw of a die will be P = 3/ 6 = ½ 

Which means, q = 1 – p = ½

Now, X has a binomial distribution therefore, P (X = x) 

(i) P (5 successes) 

(ii) P (at most 5 successes) P (x ≤ 5)

= 63/ 64

Ans. Let A be the event when the number obtained is even.

And event B be event when the number obtained is red.

P(A) = \(\frac{3}{6} = \frac{1}{2}\), P(B) = \(\frac{3}{6} = \frac{1}{2}\)

P(A∩B) = P (getting an even red number) = \(\frac{1}{6}\)

Since P(A) . P(B) = \(\frac{1}{2} . \frac{1}{2} = \frac{1}{4}\) ≠ P(A∩B) which is \(\frac{1}{6}\)

∴ A and B are not independent events.